Question: Let $V$ be a simple solid region oriented with outward normals that has a piecewise-smooth boundary surface $S$. $ \oiint_S \left[ \sin(x^2yz) \hat{\imath} + (x^2 - z^2) \hat{\jmath} + (xy + z) \hat{k} \right] \cdot dS$ Use the divergence theorem to rewrite the surface integral as a triple integral. $ \iiint_V $ $ \, dV$
Solution: Assume we have a simple solid region $V$ oriented with outward normals, and it has a piecewise-smooth, closed boundary surface $S$. If $F$ is a continuously differentiable vector field in $\mathbb{R}^3$, then the divergence theorem says: $ \oiint_S F \cdot dS = \iiint_V \text{div}(F) \, dV$ The given surface, boundary, and vector field satisfy the conditions for the divergence theorem. We're converting a surface integral into a triple integral, so we know $F$ and we want to find $\text{div}(F)$. $\begin{aligned} F(x, y, z) &= \sin(x^2yz) \hat{\imath} + (x^2 - z^2) \hat{\jmath} + (xy + z) \hat{k} \\ \\ \text{div}(F) &= \dfrac{\partial}{\partial x} \left[ \sin(x^2yz) \right] \\ \\ &+ \dfrac{\partial}{\partial y} \left[ x^2 - z^2 \right] \\ \\ &+ \dfrac{\partial}{\partial z} \left[ xy + z \right] \\ \\ &= 2xyz \cos(x^2yz) + 1 \end{aligned}$ Therefore, the equivalent triple integral is: $ \iiint_V 2xyz \cos(x^2yz) + 1 \, dV$